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\docbegin
\setsection{4}

\section{Linear Operators on Hilbert Spaces}

\begin{Exec}\label{Exec5.1}
  Let $\ch$ be a Hilbert space and let $y,z\in\ch$.
  If $T$ is the bounded linear operator defined by
  $T(x)=\langle x,y\rangle z$ for all $x\in\ch$ (cf. Exercise~4.6),
  show that $T^\ast(w)=\langle w,z\rangle y$ for all $w\in\ch$.
\end{Exec}

\begin{proof}
  Since $\inp{T(x)}{w} = \inp{\inp{x}{y}z}{w} = \inp{x}{y}\inp{z}{w}
  = \inp{x}{\overline{\inp{z}{w}}y} = \inp{x}{\inp{w}{z}y}$ for all
  $x,w\in \ch$, $T^\ast = \inp{w}{z}y$.
\end{proof}

\begin{Exec}
  Let $L^2[a,b]$ be as in Example~\ref{Exam5.1.2} 
  and let $\phi : [c, d]\times [a, b]\to\bc$ ($c,d\in\br$ with $c<d$)
  be a fixed continuous function, and consider the operator $T :
  L^2[a, b]\to L^2[c, d]$, where for every $f\in L^2[a, b]$, the
  function $Tf\in L^2[c, d]$ is defined by
  $$
  (Tf)(s) = \int_a^b \phi(s, t)\,f(t)\, dt\quad\text{for all}\ s\in
  [c, d].
  $$
  Show that $T\in \cb(L^2[a,b], L^2[c,d])$ and find $T^\ast$.
\end{Exec}

\begin{proof}
  Clearly $T$ is linear. As
  \[
  \|Tf\|^2 = \int_c^d (\int_a^b \phi(s,t)\, f(t)\, dt)^2 ds
  = \int_c^d \int_a^b |\phi(s,t)|^2 dt \int_a^b |f(t)|^2 dt ds 
  \leq M \|f\|^2
  \]
  where $M = (b-a)(d-c)\max\{\phi(s,t)^2: (s,t)\in [c,d]\times[a,b]\}<\infty$
  for $\phi$ is continuous and $[c,d]\times[a,b]$ is compact,
  $T\in \cb(L^2[a,b], L^2[c,d])$.
  
  By Fubini Theorem,  
  \[
  \inp{Tf}{g} = \int_c^d \int_a^b \phi(s, t)\,f(t)\, dt \overline{g(s)}ds
  = \int_a^b f(t) \overline{\int_c^d \overline{\phi(s,t)} g(s) ds} dt
  = \inp{f}{\int_c^d \overline{\phi(s,t)}\, g(s)\, ds}
  \]
  so that $(T^\ast g)(t) = \int_c^d \overline{\phi(s,t)}\, g(s)\, ds
  \quad\text{for all}\ t\in [a, b]$.
\end{proof}

\begin{Exec}\label{Exec5.3}
  Consider the vector space $\ell^2$ of all square summable infinite
  sequences of complex numbers, with inner product 
  $\langle x, y\rangle =\sum_{i=1}^\infty x_i\overline{y_i}$. 
  Find the adjoint
  operator $T^\ast$ of the bounded linear operators
  $T:\ell^2\to\ell^2$ defined by
  $$
  T(x) = (0, x_1, x_2, x_3, \cdots) \quad\text{for every}\ x = (x_1,
  x_2, x_3,\cdots)\in\ell^2.
  $$
\end{Exec}

\begin{solution}
  As 
  \[
  \inp{T(x)}{y} = \inp{(0, x_1, x_2, \cdots)}{(y_1, y_2,  \cdots)}
  = \sum_{i=1}^\infty x_i \overline{y_{i+1}} \quad \text{for every}\ y\in \ell^2,
  \]
  $T^\ast(y) = (y_2,y_3,\cdots)$.
\end{solution}

\begin{Exec}\label{Exec5.4}
  Find the adjoint  $T^\ast$ of the bounded linear operators 
  $T:\ell^2\to\ell^2$ defined
  by
  $$
  T(x) = (0, 4x_1, x_2, 4x_3, x_4,\cdots)\quad \text{for every} \ x =
  (x_1, x_2, x_3, x_4,\cdots)\in\ell^2
  $$
  as discussed in Exercise~4.4.
\end{Exec}

\begin{solution}
  As 
  \[
  \inp{T(x)}{y} = \inp{(0, 4x_1, x_2,\cdots)}{(y_1,y_2,y_3,\cdots)}
  = \sum_{i=1}^\infty 4x_{2i-1} \overline{y_{2i}} + x_{2i} \overline{y_{2i+1}}
  \quad \text{for every}\ y\in \ell^2,
  \]
  $T^\ast(y) = (4y_2,y_3,4y_4,y_5\cdots)$.
\end{solution}

\begin{Exec}\label{Exec5.5}
  Let $T$ be a bounded linear operator on Hilbert space $\ch$.  Show
  that if $\Re\langle Tx,x\rangle=0$ for all $x\in\ch$ then
  $T+T^\ast=0$.
\end{Exec}

\begin{proof}
  Since $\Re\inp{Tx}{x} = 0$, $\inp{(T+T^\ast)(x)}{x} = 
  \inp{Tx}{x} + \inp{T^\ast x}{x} 
  = \inp{Tx}{x} + \overline{\inp{x}{T^\ast x}}
  = 2\Re\inp{Tx}{x} = 0$ for every $x\in \ch$.
  
  We will show that $G = 0$ if and only if $\inp{Gx}{x} = 0$.
  Then $T+T^\ast = 0$.
  
  In fact, we just have to show $G=0$ if $\inp{Gx}{x}=0$ for the 
  other side is obvious.
  Since for all $x,y\in \ch$, 
  \[ 
  0 = \inp{G(x+y)}{x+y} 
  = \inp{Gx}{x} + \inp{Gx}{y} + \inp{Gy}{x} +\inp{Gy}{y}
  = \inp{Gx}{y} + \inp{Gy}{x}
  \]
  and 
  \[ 
  0 = i\inp{G(x+iy)}{x+iy} = i\inp{Gx}{iy} + i\inp{Giy}{x}
  = \inp{Gx}{y} - \inp{Gy}{x},
  \]
  $\inp{Gx}{y} =0$ so that $G=0$.
\end{proof}

\begin{Exec}\label{Exec5.6}
  Show that $T\in\cb(\ell^2)$ defined in Exercise 5.3 is not
  invertible (Hint: show that $(1,0,0,0,\cdots)\in \text{Ker}\,
  T^\ast$ and use Corollary~\ref{Cor5.1.1}).
\end{Exec}

\begin{proof}
  Clearly $T^\ast((1,0,0,\cdots)) = \bf 0$, 
  i.e $(1,0,0,\cdots)\in \text{Ker}\,T^\ast$.
  Hence $\overline{{\rm Im}\,T} = ({\rm Ker}\,T^\ast)^\bot \subsetneq \ell^2$ 
  so that $T$ is not invertible.
\end{proof}

\begin{Exec}\label{Exec5.7}
  Let $\ch$ be a Hilbert space and
  let $T\in\cb(\ch)$.  Show that
  $$
  \sigma(T+\mu I)=\sigma(T)+\mu\quad\text{for all}\ \mu\in\br.
  $$
\end{Exec}

\begin{proof}
  If $\lambda \in \sigma(T)$, $(\lambda+\mu)I - (T+\mu I) = \lambda I - T$ 
  is not invertible so that $\sigma(T)+\mu \subset \sigma(T+\mu I)$.
  On the other hand, if $\lambda + \mu \in \sigma(T + \mu I)$, $\lambda I - T
  = (\lambda +\mu)I - (T + \mu I)$ is not invertible Let $\ch$ be a Hilbert space and
let $T\in\cb(\ch)$.  Show that $\langle Tx,x\rangle \in\br$ for all
$x\in\ch$ if and only if $T$ is self-adjoint (Hint: use the
Polarization identity in Lemma~\ref{Lem3.1.1} for $\langle
Tx,y\rangle$ and $\langle Ty,x\rangle$).
  so that $\sigma(T+ \mu I) \subset \sigma(T)+\mu$.
  Thus $\sigma(T+\mu I)=\sigma(T)+\mu$ for all $\mu\in\br$.
\end{proof}

\begin{Exec}\label{Exec3.8}
   Let $\ch$ be a Hilbert space and let $y,z\in\ch$.
   Show that $T\in\cb(\ch)$ defined by $Tx=\langle x,y\rangle z$ (see
   Exercise~4.6) is compact.
\end{Exec}

\begin{proof}
  Suppose that $\{x_n\}$ is a bounded sequence in $\ch$.
  Clearly $\{\inp{x_n}{y}\}$ is bounded in $\bbf$.
  So there exits a subsequence $\{x_{n_j}\}$ such that $\{\inp{x_{n_j}}{y}$ 
  is convergence.
  Hence $\{Tx_{n_j}\}$ is convergence for $\ch$ is complete so that $T$ is compact.  
\end{proof}

\begin{Exec}\label{Exec5.9}
  Define operators
  $S,T\in\cb(\ell^2)$ by
  $$
  Sx=(0,\frac{x_1}1,\frac{x_2}2,\frac{x_3}3,\cdots),\quad
  Tx=(\frac{x_1}1,\frac{x_2}2,\frac{x_3}3,\cdots).
  $$
  Show that these operators are compact, $\sigma(S)=\{0\}$,
  $\sigma_p(S)=\emptyset$ and $\sigma(T)=\sigma_p(T)=\{0\}$.  Show
  also the Im\,$S$ is not dense in $\ell^2$, but Im\,$T$ is dense.
\end{Exec}

\begin{proof}
  By Exercise~\ref{Exec5.15} $S$,$T$ is compact
  (Let $e_n=(0,\cdot,0,1(n-th),0,\cdots)$ and $a_n=1/n$, then take
  $\hat{e}_n = e_{n+1}$ and $\hat{e}_n=e_n$ respectively).
  
  Clearly $Im(S) \subsetneq \ell^2$ and $S$ is injective 
  so that $0\in \sigma(S)$ and $0\notin \sigma_p(S)$.
  Let $a_n = 1/n$.
  Consider $\lambda I - S$ where $\lambda\in \bc\setminus\{0\}$.
  Note that $(\lambda I-S)(x) = 
  (\lambda x_1, \lambda x_2 - a_1 x_1, \lambda x_3 - a_2 x_2,\cdots)$.
  For every $y=(y_1,y_2,\cdots) \in \ell^2$ there is a unique solution of
  $Sx=y$ where $x_1 = y_1/\lambda, x_2 = (y_2+a_1 x_1)/\lambda,
  x_2 = (y_3 + a_2 x_2)/\lambda, \cdots$ and $x = (x_1,x_2,\cdots)$.
  %%Since $a_n\rightarrow 0$, $\sum_{j=1}^{n-i}$
  Hence $\lambda I-S$ is invertible so that 
  $\sigma(S) = \{0\}$ and $\sigma_p(S) = \emptyset$.
  
  Note that $(\lambda I-T)(x) = ((\lambda -a_1)x_1, (\lambda -a_2)x_2,\cdots)$.
  If there exist $n$ such that $\lambda = a_n$,
  $(\lambda I-T)(x + c e_n) = (\lambda I -T)x$ 
  so $T$ is not injective i.e. $a_n \in \sigma_p{T}$.
  If $\lambda \not in \{a_n\}$, $Tx=y$ have a unique solution
  for every $y\in \ell^2$.
  $x=(x_1, x_2,\cdots)$ where $x_i = y_i/(\lambda - a_n)$.
  Hence $\lambda I-T$ is invertible so that 
  $\sigma(T)=\sigma_p(T) = \{1,1/2,1/3,\cdots\}$.

  Have to proof the solution $x\in \ell^2$.
\end{proof}

\begin{Exec}\label{Exec5.10}
  Let $\ch$ be a Hilbert space and
  let $T\in\cb(\ch)$.  Show that $\langle Tx,x\rangle \in\br$ for all
  $x\in\ch$ if and only if $T$ is self-adjoint (Hint: use the
  Polarization identity in Lemma~\ref{Lem3.1.1} for $\langle
  Tx,y\rangle$ and $\langle Ty,x\rangle$).
\end{Exec}

\begin{proof}
  Since $\inp{Tx}{x}\in \br$, $\inp{(T-T^\ast) x}{x} 
  = \inp{Tx}{x} - \overline{\inp{Tx}{x}} = 0$.
  By the proof of Exercise~\ref{Exec5.3} and $T-T^\ast$ is self-adjoint,
  $T-T^\ast = 0$.
  Hence $\inp{Tx}{y} - \inp{x}{Ty} = \inp{Tx}{x} - \inp{T^\ast x}{y} = 
  \inp{(T-T^\ast)x}{y} = 0$ so that $T$ is self-adjoint. 
\end{proof}

\begin{Exec}\label{Exec5.11}
  Let $\ch$ be a Hilbert
  space over $\bbf$ and let $S\in\cb(\ch)$ be self-adjoint. Show that
  $S^n$ is self-adjoint for each $n\in\bn$ and conclude that
  $\|S^n\|=\|S\|^n$.
\end{Exec}

\begin{proof}
  Since $S$ is self-adjoint, $\inp{S^n x}{y} = \inp{S^{n-1} x}{Sy} =
  \inp{S^{n-2} x}{S^2 y} =\cdots = \inp{x}{S^n y}$ so that $S^n$ is self-adjoint.
  
  Clearly $\|S^n\| \leq \|S\|^n$.
  By Corollary~\ref{Cor5.2.2} there exists $\lambda \in \rho(T)$
  such that $|\lambda| = \|S\|$.
  By Lemma~\ref{Lem5.2.7} there exits a sequence $\{x_k\}$ such that 
  $\lim_{k\rightarrow \infty} \|(\lambda I - S)x_k\|=0$
  and $\|x_k\| = 1$ for every $k\in \bn$.
  As
  \[
  \begin{split}
    &\limsup_{k\rightarrow \infty} \|(\lambda^n I - S^n)x_k\| \\
    =& \limsup_{n\rightarrow \infty} 
    \|(\lambda^{n-1} I + \lambda^{n-2} S + \lambda^{n-3} S^2 + \cdots + S^{n-1})
    (\lambda I - S)x_k\|\\
    \leq& \|\sum_{i=0}^{n-1} \lambda^{n-1-i} S^i\|
    \limsup_{k\rightarrow \infty}\|(\lambda I - S) x_k\| 
    = 0,
  \end{split}
  \]
  $\|S^n\| \geq \limsup_{k\rightarrow\infty}\|S^n x_k\| =
  \limsup_{k\rightarrow \infty} \|S^nx_k\| + \|(\lambda^n I - S^n)x_k\|
  \geq  \limsup_{k\rightarrow \infty} \|\lambda^n I x_k\| = |\lambda|^n = \|S\|^n$.
  Thus $\|S^n\|=\|S\|^n$.
\end{proof}

\begin{Exec}\label{Exec5.12}
  Let $T\in\cb(\ch)$ be a self-adjoint
  operator on a complex Hilbert space $\ch$.  If $\sigma(T)$ contains
  exactly one point $\lambda$, show that $T=\lambda I$.
\end{Exec}

\begin{proof}
  Suppose that $T\neq \lambda I$. So $T-\lambda I \neq 0$, 
  then $\|T-\lambda I\|$ or $-\|T-\lambda I\|$ 
  is an element of $\rho(T-\lambda I)$.
  Since $\rho(T-\lambda I) = \rho(T) - \lambda$ by Exercise~\ref{Exec5.7},
  it follows that $\lambda + \|T-\lambda I\|$
  or $\lambda - \|T-\lambda I\|$ is an element of $\rho(T)$, a contradiction 
  with $\lambda$ is the unique element of $\rho(T)$.
\end{proof}

\begin{Exec}\label{Exec5.13}
  Show that an orthonormal sequence
  $\{e_n\}$ in a Hilbert space $\ch$ cannot have a convergent
  subsequence.
\end{Exec}

\begin{proof}
  If $\{e_n\}$ have a convergent subsequence $\{e_{n_j}\}$ such that
  $e_{n_j} \rightarrow x\in \ch$ as $j \rightarrow \infty$.
  Clearly $X = \overline{\sspan\{e_n\}}$ is a closed linear subspace of $\ch$.
  So $x \in X$.
  As $0 = \lim_{j\rightarrow \infty} \inp{e_{n_j}}{e_k} = \inp{x}{e_k}$
  for every $k\in bn$, $x \in (\sspan\{e_n\})^\bot = X^\bot$ so that $x=0$.
  On the other hand,  $\|x\| = \lim_{j\rightarrow \infty} \|e_{n_j}\| = 1$,
  a contradiction.
  Thus $\{e_n\}$ cannot have a convergent subsequence.
\end{proof}

\begin{Exec}\label{Exec5.14}
  Let $\ch$ be an
  infinite-dimensional Hilbert space with an orthonormal basis
  $\{e_n\}$ and let $T\in\cb(\ch)$.  Show that if $T$ is compact then
  $\lim_{n\to\infty}\|Te_n\|=0$.
\end{Exec}

\begin{proof}
  Clearly $0\leq \liminf_{n\rightarrow \infty} \|Te_n\|$.
  Suppose that $\limsup_{n\rightarrow \infty} > 0$. Then there exists a subsequence 
  $\{e_{n_j}\}$ and $\varepsilon >0$ 
  such that $\|Te_{n_j}\| > \varepsilon$ for every $j\in\bn$.
  Since $\{e_{n_j}\}$ is bounded and $T$ is compact
  there exists a subsequence $\{e_{n_{j_k}}\}$
  such that $Te_{n_{j_k}} \rightarrow x$ as $k \rightarrow \infty$.
  By Bessel inequality
  $\inp{x}{e_n} = \inp{\lim_{k\rightarrow \infty} Te_{n_{j_k}}}{e_n}
  = \lim_{k\rightarrow \infty} \inp{e_{n_{j_k}}}{T^\ast e_n} = 0$ for every $n\in\bn$.
  Since $\{e_n\}$ is an orthonormal basis of $\ch$, it gives that $x=0$.
  So $0=\|x\| = \lim_{k\rightarrow \infty} \|Te_{n_{j_k}}\| < \varepsilon$, 
  a contradiction with $\|Te_{n_j}\| >\varepsilon$ for every $j\in\bn$.

  Hence $\limsup_{n\rightarrow \infty}\|Te_n\| =0$ 
  so that $\lim_{n\rightarrow \infty} \|Te_n\|=0$.
\end{proof}

\begin{Exec}\label{Exec5.15}
  Let $\ch$ be an
  infinite-dimensional Hilbert space and let $\{e_n\}$,
  $\{\widehat{e}_n\}$ be orthonormal sequences in $\ch$.  Let
  $\{\alpha_n\}$ be s sequence in $\bc$ and define a linear operator
  $T:\ch\to\ch$ by $Tx=\sum_{n=1}^\infty \alpha_n\langle x,e_n\rangle
  \widehat{e}_n$.  Show that $(a)$ $T$ is bounded if and only if the
  sequence $\{\alpha_n\}$ is bounded; $(b)$ $T$ is compact if and only
  if $\lim_{n\to\infty} \alpha_n=0$; $(c)$ $y\in$ Im\,$T$ if and only
  if $y=\sum_{n=1}^\infty \alpha_n\xi_n \widehat{e}_n$ for some
  $\{\xi_n\}\in\ell^2$, and deduce that if infinitely many of
  $\alpha_n$ are nonzero and $\lim_{n\to\infty}\alpha_n=0$ then
  Im\,$T$ is not closed.
\end{Exec}

\begin{proof}
  \begin{itemize}
  \item [(a)] If $\{\alpha_n\}$ is bounded, 
    there exist $M>0$ such that $|\alpha_n| < M$ for every $n\in \bn$. 
    By the Pythagorean theorem and the Bessel inequality
    $\|Tx\|^2 = \|\sum_{n=1}^\infty \alpha_n\inp{x}{e_n}\hat{e}_n\|^2
    = \sum_{n=1}^\infty |\alpha_n|^2|\inp{x}{e_n}|^2 \leq M^2 \|x\|^2$.
    So $T$ is bounded.
    
    Suppose that  $\{\alpha_n\}$ is not bounded.
    Then there exists a subsequence $\{\alpha_{n_i}\}$
    such that $|\alpha_{n_i}|\rightarrow \infty$.
    Hence $\sup_{\|x\|=1} \|Tx\|\geq \|Te_{n_i}\| 
    = |\alpha_{n_i}| \rightarrow \infty$
    so $T$ is not bounded.
    
    Thus $T$ is bounded if and only if the
    sequence $\{\alpha_n\}$ is bounded.
  \item [(b)] If $T$ is compact, $\lim_{n\rightarrow\infty} |\alpha_n|
    = \lim_{n\rightarrow\infty} \|T_{e_n}\| = 0$ by Exercise~\ref{Exec5.15}.
    
    Suppose that $\lim_{n\to\infty} \alpha_n=0$.
    For every $\varepsilon >0$ 
    there exist $K$ such that $|\alpha_k|<\varepsilon$ for every $k>K$.
    If $\{x_n\}\in \ch$ is bounded, 
    $\|\inp{x_n}{e_i}\| < \|x_n\|$ is bounded for every $i\in \bn$.
    
  \end{itemize}
\end{proof}

\begin{Exec}\label{Exec5.16}
  Suppose that $\ch$ is a Hilbert
  space, $T\in\cb(\ch)$ is a self-adjoint compact operator on $\ch$
  and $\mathcal{N}$ is a closed linear subspace of $\ch$ which is
  invariant under $T$ (i.e. $T(\mathcal{N})\subset \mathcal{N}$).  Let
  $T|_\mathcal{N}$ denote the restriction of $T$ to $\mathcal{N}$.
  Show that $T|_\mathcal{N}$ is a self-adjoint compact operator on the
  Hilbert space $\mathcal{N}$.
\end{Exec}

\begin{proof}
  For every $x,y\in \mathcal{N}$,
  $T|_\mathcal{N}(x)=T(x), T|_\mathcal{N}(y)=T(y) \in \mathcal{N}$.
  Since $T$ is self-adjoint, $\inp{T|_\mathcal{N}(x)}{y} = \inp{Tx}{y}
  = \inp{x}{Ty} = \inp{x}{T|_\mathcal{N}(y)}$ 
  so that $T|_\mathcal{N}$ is self-adjoint.
  
  For every bounded sequence $\{x_n\}\in \mathcal{N}$, clearly $\{x_n\}$ 
  is bounded in $\ch$. So $\{Tx_n\}$ 
  have a convergent subsequence $\{Tx_{n_j}\}$ which converges to $y$ in $\ch$.
  Since $\mathcal{N}$ is closed 
  and $\{T|_\mathcal{N} x_{n_j}=Tx_{n_j}\} \in \mathcal{N}$,
  it gives that $y\in \ch$
  so that $\{T|_\mathcal{N} x_{n_j}\}$ is convergence in $\mathcal{N}$.
  Thus $T|_\mathcal{N}$ is compact.
\end{proof}

\begin{Exec}\label{Exec5.17}
  Prove Proposition~\ref{Pro5.2.1}. 
  (i.e. Let $\ch$ and $T$ be as in {\em Lemma~\ref{Lem5.2.5}}.  If
  $\mathcal{V}$ is a $T$-invariant subspace, then
  $T\in\cb(\mathcal{V}^{\bot})$ and it is self-adjoint compact.)
\end{Exec}

\begin{proof}
  Suppose that $x\in \mathcal{V}^\bot$. 
  Since $T$ is self-adjoint and
  $Ty \in \mathcal{V}$ for every $y\in \mathcal{V}$,
  $\inp{Tx}{y} = \inp{x}{Ty}$ so that $Tx \in \mathcal{V}^\bot$ 
  i.e. $T\in \cb(\mathcal{V}^{\bot})$.
  
  By Exercise~\ref{Exec5.16} $T$ is self-adjoint compact.
\end{proof}

\docend